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Let $P$ be a polynomial of degree $\geq 2.$ Consider the I.V.P. $$\frac{dy}{dx}=P(y),y(0)=1.$$ Now by Picard Uniqueness theorem the above I.V.P has unique solution in any BOUNDED interval of $\mathbb{R}$ containing $0$ as our $f(x,y)$ is Globally Lipschitz. But i want some example such that the above problem does't have unique solution(more than degree of $P$) on $\mathbb{R}.$ Please help. Thanks a lot.

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  • $\begingroup$ Have you tried $y' = y^2$? $\endgroup$ – Alex Macedo Jan 6 '17 at 16:52
  • $\begingroup$ how it has more than one solution $\endgroup$ – neelkanth Jan 6 '17 at 16:53
  • $\begingroup$ its solution is given by $\frac{-1}{x+c}$ and apply I.C. $\endgroup$ – neelkanth Jan 6 '17 at 16:53
  • $\begingroup$ The solution on $(-\infty, 1)$ is $y = 1/(1 - x)$. On $(1, \infty)$ you can give $y$ the same rule or set $y = 0$. $\endgroup$ – Alex Macedo Jan 6 '17 at 16:55
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    $\begingroup$ And even with $y' = y^2$ you have infinitely solutions, right? On $(1, \infty)$, you can set $y = 1/(x-c)$ for any $c \in \mathbf R$. $\endgroup$ – Alex Macedo Jan 6 '17 at 17:02
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A polynomial is decidedly not globally Lipschitz, it is continuously differentiable and thus locally Lipschitz, but that is all.

As a consequence, the relevant version of the existence and uniqueness theorem is that for any initial condition there exists a unique solution on a small time interval around the initial time. This germ of a solution can be extended to a maximal solution which is also unique on its domain. Note that the domain will in general depend on the initial point.

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  • $\begingroup$ But my question is about example ... $\endgroup$ – neelkanth Jan 6 '17 at 17:14
  • $\begingroup$ The same applies there. You were given examples for dynamical blow-up, make it more flexible using $y'=c(y-a)^2$. Compare with $y'=c(y-a)(y-b)$, $a<b$, which you can solve directly using separation and partial fraction decomposition. $\endgroup$ – LutzL Jan 6 '17 at 17:19

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